No you guys, most shapes do not have a center of mass with this property (but some do)

I remember thinking to myself "I'm not sure if such a point exists, but if it did then we wouldn't have needed to do all this work, so it probably can't be proven to generally exist".

but but but life would be so much easier if it WAS true!

I love how confident all the commenters were that they figured it out

Shortcut on finding the areas at 2:27: The top triangle is similar to the whole triangle, with a ratio of 2/3. So the area ratio is (2/3)^2 = 4/9 (which is not 1/2).

Thanks for shouting STEMerch EU out, Zach <3 :)

I was one of those people who commented something like this. For some reason it seems intuitive to me that a "magical point" would always exist, but my intuition was wrong. Super interesting!

Huh... So I've done a lot of simulation stuff in the past using boring filled solids as approximations. My immediate reaction to the premise was surprise, but the centroid doesn't just measure the volume, but the volume AND distance. A large lump of volume near the centroid can be balanced out by a small amount of volume far from it. Derp. Good to be humbled and remember the basics every once in a while. >_<

People who commented on other video: "this seems easy just do this, why make a video. " Those same people this episode: "I totally agree, good thing I never said that last episode" deletes comment

It's really good that you addressed this! Very often I read a neat proof but if I didn't try to solve it myself I'd have no clue why they chose to use that idea, and why other easier ones don't work.

Was wondering why there were no comments until I realised this was just posted. Darn, I can't read other people's insights until after I've watched it...

This is what happens when a core audience is comprised of engineers instead of physicists! Love your vids btw

For those who are still confused, the triangle balances across the horizontal line because the center of mass of the top triangle is farther away from the line than the center of mass of the bottom trapezoid. So long story short leverage is weird.

Thanks for pointing out I was being dumb! You made me less dumb now. I didn't comment it, but I definitely thought it. My intuition was the same as everyone else's: "If I can balance a shape on a line, there's equal mass (and thus area) on each side". Cool, now try that with a spoon.

However, a cool fact is that all convex shapes have a point in their interior whereby any lime drawn through it will cut the shape into two pieces, each with area at least 4/9ths of the shapes original area.

I think the simplest way of seeing how this doesn’t work is an L shape, when the centre of gravity is outside the shape. Or any shape with the centre of mass outside. It doesn’t cut the area into half because the centre of mass depends on the torque/moments instead of the distribution of mass.

No, quite simply, the masses don’t need to be equal the same on either side of a solid (or thin object). What needs to be same are the moments (integral *). You can have a ‘solid’ that is a 1kg, and a 10kg particle, exactly 11 m apart, and you will find that centre of mass is on the line joining them, 1m from the 10kg weight and 10m from the 1kg weight. For a simple case, consider the axis (of rotation) to be a line perpendicular to this system, of course passing through the centre of mass. One side obviously has ten times as much mass as the other, yet the moments are equal (1 kg*10 m = 10 kg*1 m).

The equilateral triangle's case is a particularly easy one to disprove using simple geometry. You can prove that there are five congruent equilateral triangles on the bottom and four on the top. No trig or irrational numbers necessary. It's quite beautiful how you can prove that 4 to 5 ratio with two different methods

A line intersecting the center of mass does not cut the shape into two parts of equal area. Instead, the invariant is going to be based on the area weighted by its position relative to the center of mass.

The moduli space of oriented planes in R^3 is homeomorphic to S^2 × R in the obvious way. This maps into I^3, whose coordinates represent the portion of mass of each object on the positive side of the plane. This map extends to the compactification S^3, where one pole at negative infinity maps to (0, 0, 0) and the other pole at positive infinity maps to (1, 1, 1). It's now a trivial exercise in topology to conclude that there is a point that maps to (1/2, 1/2, 1/2). Hint: Excising this point leaves a target space with homotopy type S^2. Hint 2: Consider what happens when you reverse the orientation of a plane.

I've actually come to this conclusion for 3d space by asking a somewhat similar question regarding fluid containers: I was looking for some rule or algorithm by which I could quickly find the half-way point where a container is half-full. I started with the known case of a cylinder, since half-way up the cylinder is half, halfway deep is half, and halfway into the breadth is half, since all those planes intersect at right angles, I figured that for all possible orientations of a half-full cylinder, the fluid level would be tangential to a single point, being the center of volume for the cylinder. I continued that line of reasoning, thinking of what the easiest way to find the half-way point in a cup was. I started by thinking of cups as cones, and immediately came to the conclusion you stated upon discovery that cones have infinite orientations where a line plane through the center of mass would have been above the half-way point in the container. My question is if there is a name for the shape that's described as the region where all planes that bisect a polyhedral area in half are tangential to this shape. Literally realized this seconds after posting this comment, but the plane that bisects all three objects will simultaneously be tangential to some part of the surface of each of their respective 'half-oids'